$\text{Đáp án:}$
$\text{Giải thích các bước giải:}$
$\text{2) a²+b²+c²+d²≥(a+b).(c+d)}$
$\text{⇔a²+b²+c²+d²≥ac+bc+bd+ad}$
$\text{⇔2a²+2.b²+2.c²+2.d²≥2ac+2bc+2bd+2ad}$
$\text{⇔(a²-2ac+c²)+(b²-2bc+c²)+(b²-2bd+d²)+(a²-2ad+d²)≥0}$
$\text{⇔(a-c)²+(b-c)²+(b-d)²+(a-d)²≥0}$
$\text{3) Xét y³-x³=x³+2x²+3x+2-x³}$
$\text{=2x²+3x+2}$
$\text{Ta chứng minh A=2x²+3x+2≥0}$
$\text{⇔8(2x²+3x+2)≥0}$
$\text{⇔16x²+24.x+16≥0}$
$\text{⇔(4x)²+2.(4x).3+9+7≥0}$
$\text{⇔(4x+3)²+7>0 với mọi x nên 2x²+3x+2>0}$
$\text{Hay y³-x³>0 (1)}$
$\text{⇔y>x}$
$\text{Mặt khác (x+3)³-y³}$
$\text{=x³+9.x²+27x+27-(x³+2.x²+3x+2)}$
$\text{=7.x²+24x+25}$
$\text{Ta chứng minh 7.x²+24x+25≥0}$
$\text{⇔(7x)²+2.(7x).12+12²+31≥0}$
$\text{⇔(7x+12)²+31≥0}$
$\text{nên (x+3)³-y³≥0}$
$\text{⇔(x+3)³>y³(2)}$
$\text{(1);(2)⇒(x+3)³>y³>x³}$
$\text{⇔x+3>y>x}$
$\text{nên y∈{x+1;x+2}}$
$\text{+) Nếu y³=(x+1)³}$
$\text{⇔x³+2x²+3x+2=x³+3.x²+3x+1}$
$\text{⇔x²=1 (rút gọn hai vế)}$
$\text{⇔x∈{1;-1}}$
$\text{⇒y∈{2;0}}$
$\text{Nên (x;y)=(1;2);(-1;0)}$
$\text{+) Nếu y³=(x+2)³}$
$\text{⇔x³+2x²+3x+2=x³+6.x²+12x+8}$
$\text{⇔4.x²+9x+6=0 (vô nghiệm)}$
